There exists $n$ such that there exists an n-digit number M for which $M(10^n + 1)$ is a perfect square. Show that there exists a positive integer $n$ greater than 3 such that  there exists an $n$-digit positive integer $M$ for which $M(10^n + 1)$ is a perfect square.
 A: As I explained in the comments, this is impossible if $10^n+1$ is squarefree. On the other hand, if $p^2$ divides $10^n+1$ for some prime $p$, then $M_0=\frac{10^n+1} {p^2}$ is an integer, and $M_0(10^n+1)=\left(\frac{10^n+1} {p}\right)^2$ is a perfect square. $M_0$ does not necessarily have $n$ digits; but we can always multiply it by some power of $4$ or $9$ in such a way that the product $M$ has $n$ digits, and then $M$ will satisfy the conditions.
So all we have to do is prove that there is some $n$ for which $10^n+1$ is divisible by the square of a prime number $p$; $p$ cannot be $2$, $3$ or $5$, so we will try $p=7$. We want to have
\begin{align}10^n \equiv -1 \pmod{49}.\end{align}
By Euler's theorem
$$10^{42}\equiv 1 \pmod{49},$$
and thus
$$(10^{21}+1)(10^{21}-1)\equiv 0 \pmod{49}.$$
To conclude that $49|10^{21}+1$ it suffices to prove that it does not divide $10^{21}-1$. We have
\begin{align}10^{21}\equiv (10^{3})^7\equiv 10^3\equiv 3^3\equiv 27\equiv -1\pmod 7, \end{align}and thus $7\not |10^{21}-1$, which implies that $49\not |10^{21}-1$.
